Using Kepler's third law, determine the period

In summary, the conversation discusses the calculation of the orbital period of a hypothetical planet located three times farther from the sun than Earth. Using Kepler's Third Law, the correct answer is determined to be 5.2 years. After realizing a mistake in the initial attempt at a solution, the conversation explores the idea of calculating the mass of the hypothetical planet using the given data, but concludes that it is not possible due to the mass-independence of Kepler's Laws. This is because the mass of the object is considered insignificant in comparison to the primary (the Sun).
  • #1
Calpalned
297
6

Homework Statement


If there were a planet three times farther from the sun than the Earth is, how long would it have taken this hypothetical planet to orbit the Sun? Assume the orbit is a circle.

Homework Equations


Kepler's 3rd Law ##= (\frac{r_1}{r_2})^3 = (\frac{T_1}{T_2})^2 ##

The Attempt at a Solution


##= (\frac{3r_E}{r_E})^3 = (\frac{T}{1})^2 ##
T = 3 years Correct answer: T = 5.2 years
Where is my blunder? Thank you all so much.
 
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  • #2
Calpalned said:

The Attempt at a Solution


##= (\frac{3r_E}{r_E})^3 = (\frac{T}{1})^2 ##
T = 3 years Correct answer: T = 5.2 years
Where is my blunder? Thank you all so much.
Show the math one step at a time.
 
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  • #3
gneill said:
Show the math one step at a time.
I just found out my error. ##3^3 = 27##. I accidentally wrote 9 instead of 27.
Thanks for helping Gniell
 
  • #4
I have one more question...
Part B of this question asks if I can use the data to solve for the mass of the hypothetical planet. The answer is no. "No mass data can be calculated from this relationship, because the relationship is mass- independent. Any object at the orbit radius of 3 times the Earth’s orbit radius would have a period of 5.2 years, regardless of its mass. "

The "...regardless of its mass" statement seems counter-intuitive to me. What if that planet turned out to be a super-giant red star much greater in mass than the Sun? Then the Sun would orbit the red star right? How can the red star still have an orbital period of 5.2 years? Thank you.
 
  • #5
The "...regardless of its mass" statement needs to be taken in the context of the basic assumption of Kepler's Laws where the masses of all the planets are considered to be much less than that of the primary (the Sun).

So I guess one might say that the mass of the object doesn't matter as long as it is insignificant :smile:
 
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  • #6
gneill said:
The "...regardless of its mass" statement needs to be taken in the context of the basic assumption of Kepler's Laws where the masses of all the planets are considered to be much less than that of the primary (the Sun).

So I guess one might say that the mass of the object doesn't matter as long as it is insignificant :smile:
Thank you
 

Related to Using Kepler's third law, determine the period

What is Kepler's third law?

Kepler's third law, also known as the law of harmonies, states that the square of the orbital period of a planet is proportional to the cube of its semi-major axis.

How is Kepler's third law used to determine the period?

By knowing the semi-major axis of a planet's orbit, which can be calculated from its distance from the sun, and using the proportionality constant, the period of the planet's orbit can be determined.

What units are used for Kepler's third law?

The units for Kepler's third law depend on the units used for the distance and period measurements. However, the most commonly used units are astronomical units (AU) for distance and years for period.

Can Kepler's third law be applied to all planets?

Yes, Kepler's third law can be applied to all planets as long as their semi-major axis and period of orbit are known. It also works for other objects orbiting a central body, such as moons orbiting a planet.

What are some limitations of using Kepler's third law to determine the period?

Kepler's third law assumes that the orbit is circular, which is not always the case. It also assumes that the mass of the central body is much greater than the orbiting object. Additionally, it does not take into account the influence of other objects in the system. These factors can result in slight variations from the predicted period.

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